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Math Help - optimzation problem #4

  1. #1
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    optimzation problem #4

    A chain of stores has been selling a line of cameras for 50 dollars each and has been averaging sales of 8000 cameras a month. They decide to increase the price, but their market research indicates that for each one dollar increase in price, sales will fall by 100.

    a) find the demand function

    b)find the price that will maximize their revenue
    Last edited by mr fantastic; August 12th 2009 at 03:43 PM. Reason: Changed post title
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  2. #2
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    Quote Originally Posted by skeske1234 View Post
    A chain of stores has been selling a line of cameras for 50 dollars each and has been averaging sales of 8000 cameras a month. They decide to increase the price, but their market research indicates that for each one dollar increase in price, sales will fall by 100.

    a) find the demand function

    b)find the price that will maximize their revenue
    let x = number of $1 price increases

    demand function ...

    d = 8000 - 100x

    revenue = demand * price ...

    R = (8000-100x)(50+x)

    find dR/dx and determine the value of x that maximizes revenue ... don't forget that the price is (50+x).
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  3. #3
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    Quote Originally Posted by skeeter View Post
    let x = number of $1 price increases

    demand function ...

    d = 8000 - 100x

    revenue = demand * price ...

    R = (8000-100x)(50+x)

    find dR/dx and determine the value of x that maximizes revenue ... don't forget that the price is (50+x).
    ok.. question, your demand function is d=8000-100x
    The answer they have is..
    p(x)=130-x/(3000)
    and b) they have $9.50.......
    I got 50+15=65 dollars with the demand function i used from above(yours: d=8000-100x)

    I am not sure how they got their answers..
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  4. #4
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    I get a price of $65 ($15 increase) to maximize revenue also.

    No idea what their p(x) is.
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